No.617 (SKB)

skb-photoNo.617 
S. K. Balasubramanian 
(India)

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Original Problems, Julia’s Fairies – 2014 (III): September – December

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Please send your original fairy problems to: julia@juliasfairies.com


No.617 by S. K. Balasubramanian – Nice miniature showing an interesting combination of Back-to-back and Madrasi! (JV)


Definition:

Back-To-Back: When pieces of opposite colors stand back-to-back with each other on the same file (white piece is on the top of black!), they exchange their roles. A pawn on the first rank cannot move. Any piece can make an en passant capture when it has got a role of Pawn by Back-To-Back.

Madrasi: Units, other than Kings, are paralysed when they attack each other. Paralysed units cannot move, capture or give check, their only power being that of causing paralysis.


No.617 S. K. Balasubramanian
India

original – 07.10.2014
Dedicated to Mr. Pierre Tritten

Solutions: (click to show/hide)

White kf8 rb3 bd1 Black kh8 ra4 bg6 ph2

h#2             2 solutions           (3+4)
Back-To-Back
Madrasi


23 Responses to No.617 (SKB)

  1. shankar ram says:

    Nice echoed play..
    The logic of the WBh5 paralysing the BBf5.. and the WRb2 paralysing BRa1 is really interesting!

  2. Nikola Predrag says:

    Yes, a miniature with a much, much more complex content than it might appear at first glance. A perfect mechanism of four active pieces. Fifth piece of the mechanism is the passive bK, a target which functions as a switch, determining the complex O/D exchange of functions in the mechanism.

    Actually, the second glance is perhaps still not enough. It deserves to be thoroughly analyzed.

  3. Nikola Predrag says:

    Beside the reciprocal separate functions, there’s a cyclic shift in the work of whole mechanism which realizes the complete logic chain of the solution.
    A poetry.

    Piece A paralyzes B which empowers C to paralyze D (so D cannot play to paralyze C)

    I – wR=A, bR=B, wB=C, bB=D
    II – wR=C, bR=D, wB=A, bB=B

    Winchloe probably treats that combination differently than Popeye. But both programs seem to treat it in a “schizofrenic” way.

    • S.K.BalasubramanianS. K. Balasubramanian says:

      Dear Nikola ,

      Thanks for your fine explanation.

      — Bala

    • shankar ram says:

      Nikola ..
      this is a ABCD -CDAB pattern..
      which is considered to be a double reciprocal change.. not a cyclic shift..
      I tried rearranging the moves.. but no dice..

  4. seetharaman says:

    This is a complex mix of two fairy conditions! From an apparently very light force Bala has produced a beautiful problem. Well done Bala! Full credit to Nikola for finding the cyclic shift of empowering functions between the two phases! (obviously not noticed by composer).

  5. Kjell Widlert says:

    An interesting distinction: Bh5 in a) MOVES like a rook (so it observes Bf5) but it still IS a bishop (so Bf5 is paralysed (of course Bh5 is not, as the observation is not mutual)). Analogously, Rb2 in b) MOVES like a Bishop (to observe Ra1) but still IS a rook (to paralyze Ra1).
    The reasoning seems logical to me.

  6. Nikola Predrag says:

    Dear Bala, I’ve had a great pleasure thanks to your problem.
    It was clear that more then the pieces of one reciprocal pair are related with the pieces of the other pair.

    It was also very probable that all these relations should make the entire logic chain of the solution with the cyclic shift between the phases.
    Then I had to go but when I returned it was actually very easy to find the proper logic chain which presents that cycle.

    Such cyclic exchanges, in the mechanism which presents the same logic chain in both phases, create the “added value”.
    One logic, one mechanism but two phases which link the chains into a closed cycle. This creates a new “dimension”, which is not present in every chess problem.

    • S.K.BalasubramanianS. K. Balasubramanian says:

      Actually, I should thank Mr. Pierre Tritten (and WinChloe) for achieving this interesting composition. Initially I thought that a bishop, empowered to move like a rook, will paralyse only a rook and not a bishop. But, when I composed a problem and tested with Popeye 4.69, I got some erratic solutions. Then I could get that Popeye has some bugs for this combination. Then I requested Pierre to help me testing this combination by WinChloe, by sending some basic positions (not a problem) .He readily tested them and Winchloe gave different interpretation of BTB+Madrasi. Then he composed a problem with this combination (probably he is yet to publish it) where he had shown mates by bishop as bishop and by rook as rook only. I thought why should I not try for the bishop to mate as a rook and the rook to mate as a bishop, which would be slightly difficult, as the black rear piece can move away to nullify the power temporarily gained by the front white piece. The result was No.617.

      So my sincere thanks to Mr. Pierre and in fact his finding helped me to compose this problem!

  7. Nikola Predrag says:

    Hm, if four links are connected, it’s a chain. If the last link is connected with the first link, the chain is closed into a cycle.

    Double reciprocal change might be AB,CD – CD,AB.
    There are two short chains AB and CD but neither is B connected with C, nor is D connected with A.
    There’s no 4-links chain. The mere sequence of moves ABCD looks as a 4-links chain but it is directly required by the rules in any h#2 just as in any ser-#4 or PG in 2.0 moves.

    But in this problem, it’s not about the order of moves but about the chain of static functions in the mates. The potential of one piece requires the potential of the other piece etc..
    Paralyzing potential of bBf5 requires the Rook-potential of wBh5 which requires the BackToBack potential of bRh4 which requires the paralyzing potential of wRb4.

    And that makes a 4-links chain required by the idea and not by the mandatory chain B1-W1-B2-W2.

  8. Nikola Predrag says:

    By the way, the given definition of Madrasi is anything but correct.
    “…Units, other than Kings, are paralysed when they attack each other…”

    It should be “A unit (other than King) is paralysed when it’s attacked by a unit of the same type…”

    • shankar ram says:

      Right.. This definition was corrected within the 1st two years after Madrasi got invented. In the case of Grasshoppers and Chinese pieces, the attack can be non-mutual.
      Btw, the word “observed” was also used instead of attacked (maybe less violent? ;-)).

      • Nicolas Dupontdupont says:

        I see a tiny difference between “observed” and “attacked”. The latter means “can be captured” (via a legal move), while observation is by no way related to legality. Am I wrong? If not, “observed” is the correct word, at least it is the notion used by solving programs (i.e. possible legal capture is not mandatory).

      • Nikola Predrag says:

        And in this problem, there are no mutual attacks, bRh4 does not attack wRb4 and bBf5 does not attack wBh5; bBb1 does not attack wBc2 and bRa1 does not attack wRb2!
        Only wB/wR attack, either by their original power or by the power given by BackToBack, respectively.

        A possibility of en-passant paralyzes only the Pawn which has just made the double-step.

  9. Nikola Predrag says:

    Terminology should be standardized wherever it could be commonly accepted.
    I’s a bit tricky to define a general meaning of some term. Squares or pieces could be attacked, threatened, guarded, observed etc..
    Some general “hypothetical cases” should be defined/described, before defining a term. But there’s hardly any public curiosity about the standardization.

    So shortly, just in principle.
    Since “guard” is traditionally used (e.g. “flightguard”), this term might be taken as referential.

    Consider an orthodox position, wKc8,wRb7,bKa8,bBa6,bSc6.
    wR guards the squares a7,b8,c7 and wK guards b8,c7,b7. But the white pieces do not attack these squares. An attack requires at least a “hypothetically legal” possibility of capture.
    A square is attacked if a piece on that square is attacked.
    After the legal 1.Sc6-b8, b8 would not be attacked, neither by wK nor by wR.
    1.Ka8-b8 is illegal because bK would be “hypothetically” attacked by both white pieces.
    “Hypothetically legal” is used here because nothing could be actually legal after the illegal 1.Ka8-b8.
    However, bKa8 guards a7,b8,b7. So, bSb8 would be “observed” by both Black and White (but not attacked). The same would be with wSb8.
    “Observed” would mean that a piece is not necessarily attacked but occupies a square which is guarded by some piece(s).
    Perhaps Madrasi should be defined by “observing”.
    Patrol chess might be – a piece guards some square only if that piece is itself observed by some “friendly” piece.

    Fairies create various situations and every orthodox principle must be properly adjusted before the generalization.

    • Nicolas Dupontdupont says:

      So you agree with my distinction above between “observing” and “attacking”.

      I noticed a rather strange fact: consider the position wPd2d3 and bPe4, with conditions Immun and Madrasi.

      For Popeye, white is stalemated. It means that this program considers wPd3 to be Immun-observed by bPe4, although the capture e4xd3 is Immun-illegal!

      For WinChloe, white may play d3-d4 or d3xe4. It means that this program no more considers wPd3 to be Immun-observed by bPe4.

      To my mind this is WinChloe which is right.

  10. Nikola Predrag says:

    I certainly agree with a distinction but more precise definitions are desirable.
    My earlier speculations would mean that in your example, the empty square d3 would be guarded by bPe4. But wPd3 is not attacked and perhaps not even observed, which would mean that if d3 is occupied, it is not guarded.
    But I would have to more than just guess the rules of “Immun”, before speculating.

    • Nicolas Dupontdupont says:

      Immun “a capture is illegal if the (circe) rebirth square of the captured piece is occupied”.

      The same dilemma arises with T&M instead of Immun: with wPd3d4 and bPe4, is wPd3 observed by bPe4?

      The point is that the “take” part e4xd3 doesn’t admit any possible “make” part. So my feeling is that the answer to the above question should be “no”.

      I think there is no urgency to provide a full fairy definition of observation, as this is the last step of the whole process. Before that, the community (or at least a big part of it) should share the same opinion on “borderline” examples (as those I just provided), and then an appropriate definition will automatically follow.

  11. S.K.BalasubramanianS. K. Balasubramanian says:

    I am surprised to see that my simple problem has created such a long dialogue. I do not know the original definition of Madrasi, but in my opinion the Madrasi should mean “when a piece other thank king observes a piece of opposite colour, but of the same kind, then the observed piece is paralysed by the observing piece and the piece which is paralysed cannot move or guard any square and it simply occupies the square where it is sitting. however, it can paralyse its opponent counterpart.”

    Further, I strongly believe that both the wPd3 & bPe4 are mutually paralysed as they mutually observe each other. I feel that observing means hypothetically capturing.

    —- Bala

    • Nicolas Dupontdupont says:

      Ok, but what is the precise meaning of “hypothetically capturing”?

      In my Immun example, may bPe4 “hypothetically capture” wPd3, although d2 is occupied and hence “real capture” is Immun-illegal?

      In my T&M example, may bPe4 “hypothetically capture” wPd3, although the “take” half-move e4xd3 doesn’t admit any “make” part, and hence is T&M-illegal?

      Strangely enough, the 2 main solving programs differ on the answers (or maybe Popeye is buggy here, I’m gonna ask to Thomas), which is enough to make those questions interesting…

  12. Nikola Predrag says:

    I certainly don’t expect to provide “a full fairy definition” for anything, but a partial standardization for the most common cases could be achieved. At least the “observing” should have the same essence in case of Madrasi and Patrol, well, if possible.

    In your Immun-example, it is crucial to know what means “observed”, if that term is used to define Madrasi. However, if the term “attacked” is used, then this term must be defined. It’s absurd to say that two pieces paralyze each other because they attack each other. If they are paralyzed, they can’t capture each other, so there’s no mutual attack.

    That’s why some “hypothetical test-cases” should be introduced. But it requires time to write about it sufficiently clearly.

  13. S.K.BalasubramanianS. K. Balasubramanian says:

    I agree with Nikola. If the word attacked is used then after paralysis they cannot attack any more and hence they should lose the power of paralysis also. But the pieces can observe each other even after paralysis. So, I think the word ‘OBSERVED’ is more appropriate for Madrasi. Further as NSR pointed out when pieces like grass-hoppers are used, then one piece may observe the other, and the reciprocal may or may not be true. That is why I used the word observed piece is paralysed.

    I missed the point in DUPONT’s example that the other fairy condition is T&M where he asked whether bPe4 observes wPd3 when d2 is occupied in a T&M+Madrasi problem. It is really debatable. May be some experts can give better answer.

  14. Kjell Widlert says:

    The definition of Madrasi and how it should work in various fairy forms was also discussed in a long thread on MatPlus.Net in 2008 (reprinted in Mat Plus 41, July 2011). Apparently it was also discussed in Problem Paradise in 2007. But I’m sorry to say that no universal agreement was reached.

    The original Madrasi definition by Karwatkar spoke about paralysis of pieces mutually attacking each other. After some discussions in feenschach, the definition was revised to include non-mutual paralysis in the en-passant case (and later by extension to grasshoppers etc). Fairly soon – I don’t remember how soon – the term “attack” began to be replaced by “observe”. This is a much better term, because a pinned piece does not normally attack an opposing piece of the same nature (it cannot capture it) but it may well observe it – and Karwatkar too wanted a pinned piece to be able to paralyse (unless I am completely mistaken).

    Now the term “observe” must be defined, of course. A natural definition would be to say that piece A is paralysed by piece B of the same nature if A would be in check from B if it were royal (a king, for example). This works well in orthodox, with most fairy pieces and with many fairy conditions.

    Trouble comes when some other fairy condition makes the capture of A by B illegal, so that a royal piece A would not be in check. There are many examples: AntiCirce and Strict Circe(ImmuneChess) when the rebirth square is occupied; Grid Chess when A and B are in the same cell; UltraMax and UltraMin when the captures are not the longest/shortest moves available; Monochrome when A and B stand on squares of different colours; Köko when A has no neighbour; etc.

    I think there are two main interpretations here: (a) if a royal piece A would not be in check, A is not Madrasi-paralysed; (b) if a royal piece A would be in check disregarding all other fairy conditions, it is Madrasi-paralysed.
    WinChloe uses interpretation (a), as far as I know.
    Popeye mostly uses interpretation (b), but – according to Thomas Maeder in the 2008 conversation – with some exceptions: Grid Chess, Monochrome, Edgemover, Köko and Geneva Chess.
    My preference is for the simpler interpretation (a).

    Observation of friendly pieces, such as in Patrol Chess, would then be interpreted in a similar fashion: piece A is observed by piece B of the same colour, if A would be in check from B if it were a royal piece of the opposite colour. But perhaps it feels unnatural to say that wPa6 is not observed by wRa1 in AntiCirce if h1 is occupied?

    There are other interesting questions of definition with Madrasi, such as “when are two pieces considered to be of the same nature”? What if one of them is neutral? A Chameleon? That’s a story for another day.

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